Let me begin with a remark related to the question of whether . We showed that this is the case if for some , or if is Dedekind-finite.

Theorem.The axiom of choice is equivalent to the statement that any Dedekind-infinite cardinal is a square.

Proof. Let be a set. Assuming that every D-infinite cardinal is a square, we need to show that is well-orderable. We may assume that . Otherwise, replace with . Let . Assume that is a square, say . Then . By Homework problem 2, , so for some , and .

Lemma.Suppose are D-infinite sets and is an (infinite) initial ordinal. If then either or .

Proof. Let be an injection. If there is some such that we are done, so we may assume that for all there is some such that . Letting be the least such , the map is an injection of into .

By the lemma, it must be that either or else . The former is impossible since , so is well-orderable, and thus so is , and since , then is well-orderable as well.

1. We all know how to solve a linear equation such as , namely (assuming ; if then either and any is a solution, or else there are no solutions). This was known to Babylonian and Persian mathematicians (with the usual caveats about the signs of , since the notion of negative numbers had not been introduced yet.)

This is trivial, but there is a subtle point here:

Some equations have no solutions.

If we are interested in solving polynomial equations in general, at some point we will need an argument justifying that we can. For now, let us proceed formally, assuming that we will always find solutions.

Just as with linear equations, we all know as well how to solve quadratics, such as . Namely, we can factor out (if we are in the linear case, so let’s assume that this is not the case) and then complete the square. We get

so iff , or

the well known quadratic formula.

Another small subtlety appears here, namely, there is some inherent ambiguity in the meaning of the expression . We usually resolve this by “choosing a sign” of the square root. As long as we are looking at quadratic polynomials with integer (or rational, or real, or even complex) coefficients, there is a standard way of making this choice. In more general situations (in arbitrary fields) there is no such standard procedure.

Besides this subtlety, a more serious one needs to be faced. Nowadays, we are used to working with complex numbers, so the view of a square root of a negative number does not cause confusion, but this was a serious issue for many centuries, and when complex numbers were first used, many were skeptical of whether they actually made sense. It wasn’t until Gauss’ presentation of complex numbers as pairs of reals that their use became mainstream. This is related to the question of whether one can always solve an equation. The answer was “no” until complex numbers were introduced and accepted, and then it became “yes.”

2. Cubic equations. The history of the solutions to cubic and quartic equations is full of melodrama, the historical note at the end of Chapter 2 of the book is well worth reading, see also this.

[Updated December 3. The previous proof that there is a canonical bijection for all infinite ordinals was seriously flawed. Thanks to Lorenzo Traldi for pointing out the problem.]

5. Specker’s lemma.

This result comes from Ernst Specker, Verallgemeinerte Kontinuumshypothese und Auswahlaxiom, Archiv der Mathematik 5 (1954), 332-337. I follow Akihiro Kanamori, David Pincus, Does GCH imply AC locally?, in Paul Erdős and his mathematics, II (Budapest, 1999), Bolyai Soc. Math. Stud., 11, János Bolyai Math. Soc., Budapest, (2002), 413-426 in the presentation of this and the following result. The Kanamori-Pincus paper, to which we will return next lecture, has several interesting problems, results, and historical remarks, and I recommend it. It can be found here.

This set is due February 6 at the beginning of lecture. Consult the syllabus for details on the homework policy.

1. a. Complete the proof by induction that if are integers and , then for all integers .

b. This result allows us to give a nice proof of the following fact: Let be a natural number and let be a positive integer. If the -th root of , , is rational, then it is in fact an integer. (The book gives a proof of a weaker fact.) Prove this result as follows: First verify that if and , then . Show that any fraction with integers, can be reduced so . Assume that is rational, say . Then also . Express this last fraction as a rational number in terms of . Use that for all and the general remarks mentioned above, to show that is in fact an integer.

2. Show by induction that for all integers there is a polynomial with rational coefficients, of degree and leading coefficient , such that for all integers , we have . There are many ways to prove this result. Here is one possible suggestion: Consider .

3. Euclidean algorithm. We can compute the gcd of two integers , not both zero, as follows; this method comes from Euclid and is probably the earliest recorded algorithm. Fist, we may assume that are positive, since , and also we may assume that , so . Now define a sequence of natural numbers as follows:

, .

Given , if , then .

Otherwise, , and we can use the division algorithm to find unique integers with such that . Set .

Let be the set of those that are strictly positive. This set has a least element, say . By the way the algorithm is designed, this means that .

Show that , and that we can find from the algorithm, integers such that .

(If the description above confuses you, it may be useful to see the example in the book.)

4. Assume that the application of the algorithm, starting with positive integers , takes steps. [For example, if and , the algorithm gives:

Step 1: , so .

Step 2: , so .

Step 3:, so , and . Here, ]

Show that , where the numbers are the Fibonacci numbers, see Exercises 15-22 in Chapter 1 of the book.

5. Extra credit problem. With as in the previous exercise, let be the number of digits of (written in base 10). Show that

In the more general setting of rings, of which the integers are an example, it is customary to call a number prime when it satisfies the second proposition, and to call it irreducible when it satisfies the definition above. Both notions coincide for the integers, but there are examples of rings where there are irreducible elements that are not prime. We will introduce rings later on in the course.

Mathematical induction. Let . Let be a statement about integers, so for each integer , may or may not hold. Assume

holds, and

if holds then holds

Then holds.

Strong induction. Let . Let be a statement about integers. Assume

if with it is the case that holds, then also holds

Then holds.

Both induction and strong induction are consequences of the well-ordering principle. In fact, all three statements are equivalent. Most properties about the natural numbers are established by inductive arguments. Here are two examples:

Example 1. For all , .

We also have and . In fact, for all , there is a polynomial with rational coefficients, of degree and leading coefficient such that for all , .

There are a few additional remarks on the Schröder-Bernstein theorem worth mentioning. I will expand on some of them later, in the context of descriptive set theory.

The dual Schröder-Bernstein theorem (dual S-B)is the statement“Whenever are sets and there are surjections from onto and from onto then there is a bijection between and .”

* This follows from the axiom of choice. In fact, is equivalent to: Any surjective function admits a right inverse. So the dual S-B follows from choice and the S-B theorem.

* The proofs of S-B actually show that if one has injections and , then one has a bijection contained in . So the argument above gives the same strengthened version of the dual S-B. Actually, over , this strengthened version implies choice. This is in Bernhard Banaschewski, Gregory H. Moore, The dual Cantor-Bernstein theorem and the partition principle, Notre Dame J. Formal Logic 31 (3), (1990), 375-381.

* If is onto, then there is 1-1, so if there are surjections in both directions between and , then and have the same size. Of course, this is possible even if and do not.

Open question.()Does the dual Schröder-Bernstein theorem imply the axiom of choice?

The only reference I know for precisely these matters is the handbook chapter MR2768702. Koellner, Peter; Woodin, W. Hugh. Large cardinals from determinacy. In Handbook of set theory. Vols. 1, 2, 3, 1951–2119, Springer, Dordrecht, 2010. (Particularly, section 7.) For closely related topics, see also the work of Yong Cheng (and of Cheng and Schindler) on Harr […]

As other answers point out, yes, one needs choice. The popular/natural examples of models of ZF+DC where all sets of reals are measurable are models of determinacy, and Solovay's model. They are related in deep ways, actually, through large cardinals. (Under enough large cardinals, $L({\mathbb R})$ of $V$ is a model of determinacy and (something stronge […]

Throughout the question, we only consider primes of the form $3k+1$. A reference for cubic reciprocity is Ireland & Rosen's A Classical Introduction to Modern Number Theory. How can I count the relative density of those $p$ (of the form $3k+1$) such that the equation $2=3x^3$ has no solutions modulo $p$? Really, even pointers on how to say anything […]

(1) Patrick Dehornoy gave a nice talk at the Séminaire Bourbaki explaining Hugh Woodin's approach. It omits many technical details, so you may want to look at it before looking again at the Notices papers. I think looking at those slides and then at the Notices articles gives a reasonable picture of what the approach is and what kind of problems remain […]

It is not possible to provide an explicit expression for a non-linear solution. The reason is that (it is a folklore result that) an additive $f:{\mathbb R}\to{\mathbb R}$ is linear iff it is measurable. (This result can be found in a variety of places, it is a standard exercise in measure theory books. As of this writing, there is a short proof here (Intern […]

The usual definition of a series of nonnegative terms is as the supremum of the sums over finite subsets of the index set, $$\sum_{i\in I} x_i=\sup\biggl\{\sum_{j\in J}x_j:J\subseteq I\mbox{ is finite}\biggr\}.$$ (Note this definition does not quite work in general for series of positive and negative terms.) The point then is that is $a< x

The result was proved by Kenneth J. Falconer. The reference is MR0629593 (82m:05031). Falconer, K. J. The realization of distances in measurable subsets covering $R^n$. J. Combin. Theory Ser. A 31 (1981), no. 2, 184–189. The argument is relatively simple, you need a decent understanding of the Lebesgue density theorem, and some basic properties of Lebesgue m […]

Given a class $S$, to say that it can be proper means that it is consistent (with the axioms under consideration) that $S$ is a proper class, that is, there is a model $M$ of these axioms such that the interpretation $S^M$ of $S$ in $M$ is a proper class in the sense of $M$. It does not mean that $S$ is always a proper class. In fact, it could also be consis […]

As the other answers point out, the question is imprecise because of its use of the undefined notion of "the standard model" of set theory. Indeed, if I were to encounter this phrase, I would think of two possible interpretations: The author actually meant "the minimal standard model of set theory", that is, $L_\Omega$ where $\Omega$ is e […]